4.4: Inverse Functions
- Page ID
- 528
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An inverse function is a function that undoes another function: If an input \(x\) into the function \(f\) produces an output \(y\), then putting \(y\) into the inverse function \(g\) produces the output \(x\), and vice versa.
Let \(f(x)\) be a 1-1 function then \(g(x)\) is an inverse function of \(f(x)\) if
\[ f(g(x)) = g(f(x)) = x. \nonumber \]
For
\[ f(x) = 2x - 1 \nonumber \]
\[ f^{ -1}(x) = \dfrac{1}{2} x +\dfrac{1}{2} \nonumber \]
since
\[ f(f^{ -1}(x) ) = 2 \left( \dfrac{1}{2} x +\dfrac{1}{2} \right) - 1 = x \nonumber \]
and
\[ f ^{-1}(f(x)) = \dfrac{1}{2} [2x - 1] + \dfrac{1}{2} = x. \nonumber \]
The Horizontal Line Test and Roll's Theorem
Note that if \(f(x)\) is differentiable and the horizontal line test fails then
\[ f(a) = f(b) \nonumber \]
and Rolls theorem implies that there is a \(c\) such that
\[ f '(c) = 0. \nonumber \]
A partial converse is also true:
If \(f\) is differentiable and \(f '(x)\) is always non-negative (or always non-positive) then \(f(x)\) has an inverse.
\[ f(x) = x^3 + x - 4 \nonumber \]
has an inverse since
\[f'(x) = 3x^2 + 1 \nonumber \]
which is always positive.
Continuity and Differentiability of the Inverse Function
- \(f\) continuous implies that \(f^{ -1}\) is continuous.
- \(f\) increasing implies that \(f^{ -1}\) is increasing.
- \(f\) decreasing implies that \(f^{ -1}\) is decreasing.
- \(f\) differentiable at \(c\) and \(f '(c) \neq 0\) implies that \(f^{ -1}\) is differentiable at \(f (c)\).
- If \(g(x)\) is the inverse of the differentiable \(f(x)\) then
\[ g'(x) = \dfrac{1}{f '(g(x))}. \lable{Rolls} \nonumber \]
if \(f '(g(x)) \neq 0\).
Since
\[ f (g(x)) = x \nonumber \]
we differentiate implicitly:
\[\dfrac{d}{dx} f(g(x)) = \dfrac{d}{dx} x.\nonumber \]
Using the chain rule
\[ y =f(u), u = g(x)\nonumber \]
\[ \begin{align*} \dfrac{dy}{x} &= \dfrac{dy}{dy} \dfrac{du}{dx} \nonumber \\[4pt] &= f '(u) g'(x) = f '(g(x)) g'(x). \end{align*} \]
So that
\[ f '(g(x)) g'(x) = 1. \nonumber \]
Dividing, we get Equation \ref{Rolls}:
\[g'(x) = \dfrac{1}{f'(g(x))}.\nonumber \]
\(\square\)
For \(x > 0\), let
\[ f(x) = x^2\nonumber \]
and
\[ g(x) = \sqrt{x}\nonumber \]
be its inverse, then
\[ g'(x) = \dfrac{1}{2\sqrt{x}}.\nonumber \]
Note that
\[ \dfrac{d}{dx} \sqrt{x} = \dfrac{d}{dx} x^{\frac{1}{2}} = \dfrac{1}{2} x^{-\frac{1}{2}} = \dfrac{1}{2\sqrt{x}}.\nonumber \]
1. Let
\[ f(x) = x^3 + x - 4.\nonumber \]
Find
\[ \dfrac{d}{dx} f^{-1}(-4). \nonumber \]
2. Let
\[f(x) =\int_2^x \dfrac{1}{1+x^3} dx\nonumber \]
Find
\[ \dfrac{d}{dx} f^{-1}(0). \nonumber \]
Outside Links
- http://en.Wikipedia.org/wiki/Rolle%27s_theorem
- Larry Green (Lake Tahoe Community College)
Integrated by Justin Marshall.